Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# 13 6 the mole fraction of benzene in the cumulative

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Unformatted text preview: : Time for the mole fraction of A in the instantaneous distillate to fall to 0.40. Analysis: Base the calculations on A, which is the more volatile component. Initially, the distillate mole fraction in equilibrium with the initial charge is obtained from Eq. (4-8) for constant relative volatility, α: y= αx 2.5(0.30) = = 0.5172 1 + x (α − 1) 1 + 0.30(2.5 − 1) (1) When the mole fraction of A in the instantaneous distillate = yD = 0.40, the mole fraction of A in the liquid in the still is obtained form a rearrangement of Eq. (1), x= 0.40 y = = 0.2105 α + y (1 − α ) 2.5 + 0.40(1 − 2.5) Modify Eq. (13-1) to include the constant feed added to the still, noting that, with the above assumptions, W = total moles in the still = constant = W0 = 20 lbmole, dW/dt = 0, and the distillate rate, call it D = molar feed rate, F: Fx F − W dx = Dy D = Fy D dt Rearranging, dx F x F − y D = dt W0 (2) Eq. (2) is integrated as follows, noting that yD = y. Substituting Eq. (1) into Eq. (2) gives, F 10 dt = dt = W0 20 xF − t 0.2105 0 0.305 0.5 dt = 0.50t = dx dx (1 + 15x )dx . = = 2.5x αx 0.30 − 2.05x 0.30 − 1 + x (α − 1) 1 + 15x . 1 + 15x )dx . = 0.585 0.305 − 2.05x Therefore, t = 0.585/0.5 = 1.17 h Exercise 13.9 Subject: Batch distillation of a mixture of isopropyl alcohol (P) and water (W) in a column with 2 equilibrium stages and a reboiler, under conditions of a constant reflux ratio. Given: Feed contains 40 mol% P and 60 mol% W. Distillation at 1 atm with a reflux of L/V = 0.9. Vapor-liquid equilibrium data in Exercise 13.2. Assumptions: Perfect mixing in the still. No holdup on the stages or in the condenser. Find: Compositions of the residue in the still and cumulative distillate when 70 mol% of the charge has been distilled (W/W0 = 0.30). Analysis: Make calculations in terms of P, the more volatile component. Eq. (13-2) applies, where yD is the mole fraction of P in the vapor leaving the top stage, and xW is the mole fraction of P in the liquid in the reboiler. ln W0 1 = ln = 1.204 = W 0.3 0.40 xW dxW y D − xW (1) The relationship between yD and xW is obtained from a McCabe-Thiele diagram by drawing a series of operating lines of slope = L/V = 0.9. For each operating line, starting from the intersection with the 45o line, which is yD, 3 stages are stepped off to determine the corresponding xW. A typical construction that starts from yD = 0.65 is shown on the next page, where the xW = 0.37. Other sets of values are given in the following table, which also includes values of the integrand, f = 1/(yD - xW), from which the integral is evaluated by the trapezoidal rule with variable increments of ∆xW. For example, the increment of the integral from x1 to x2 is (x1 - x2)(f1/2 + f2/2). The increments are summed over the region from xW = 0.40 to the value corresponding to 70 mol% distilled. xW yD from McCabe-Thiele f = 1/(yD - xW) 0.400 0.370 0.260 0.170 0.080 0.015 0.008 0.002 0.66 0.65 0.645 0.64 0.63 0.60 0.55 0.50 3.90 3.61 2.60 2.13 1.82 1.71 1.85 2.01 Increment of Int...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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